--- title: "Unbalanced source feeding star load" subject: "Power System Analysis II" chapter: 4 key: "semester_5/power_system_analysis_ii:unbalanced-source-star-load-components" source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/unbalanced-source-star-load-components site: BCT Engineering --- # Unbalanced source feeding star load Power System Analysis II · Chapter 4: Unbalanced System Analysis - Asked in: 2076 Chaitra, 8 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/unbalanced-source-star-load-components - Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-4 ## Question A 3-phase unbalanced source with voltage given by Vabc = [200∠25° 100∠−155° 80∠100°]ᵀ V is feeding a 3-phase star connected load with series impedance per phase of Zs = 8 + j24 Ohms and load neutral impedance of Zn = j1.5 Ohms. Determine the: (i) Symmetrical components of voltage (ii) Symmetrical components of currents ## Short answer Sequence voltages from A⁻¹Vp: V0=47.77∠57.63°, V1=112.78∠-0.03°, V2=61.62∠45.88° V. Load sequence impedances: Z1=Z2=8+j24 Ω, Z0=8+j28.5 Ω. Sequence currents: I0=1.614∠-16.69°, I1=4.458∠-71.60°, I2=2.436∠-25.68° A. ## Answer Find the sequence voltages with $V_s = A^{-1}V_p$, then divide each by its own sequence impedance of the load, since a balanced star load has a diagonal sequence impedance matrix. **Assumption:** the given voltages are the phase-to-ground voltages applied to the load terminals (source neutral grounded), $a = 1\angle120^\circ$. #### Phase voltages in rectangular form $$ \begin{aligned} V_a &= 200\angle25^\circ = 181.26 + j84.52\ \text{V} \\ V_b &= 100\angle{-155^\circ} = -90.63 - j42.26\ \text{V} \\ V_c &= 80\angle100^\circ = -13.89 + j78.78\ \text{V} \end{aligned} $$ #### (i) Symmetrical components of voltage $$ \begin{aligned} V_{a0} &= \tfrac13(V_a + V_b + V_c) = 25.58 + j40.35 = 47.77\angle57.63^\circ\ \text{V} \\ V_{a1} &= \tfrac13(V_a + aV_b + a^2V_c) = 112.78 - j0.07 = 112.78\angle{-0.03^\circ}\ \text{V} \\ V_{a2} &= \tfrac13(V_a + a^2V_b + aV_c) = 42.90 + j44.24 = 61.62\angle45.88^\circ\ \text{V} \end{aligned} $$ Check: $V_{a0} + V_{a1} + V_{a2} = 181.26 + j84.52 = V_a$. #### Sequence impedances of the load $$ \begin{aligned} Z_1 = Z_2 &= Z_s = 8 + j24 = 25.30\angle71.57^\circ\ \Omega \\ Z_0 &= Z_s + 3Z_n = 8 + j(24 + 4.5) = 8 + j28.5 = 29.60\angle74.32^\circ\ \Omega \end{aligned} $$ #### (ii) Symmetrical components of current $$ \begin{aligned} I_{a0} &= \frac{V_{a0}}{Z_0} = \frac{47.77\angle57.63^\circ}{29.60\angle74.32^\circ} = 1.614\angle{-16.69^\circ}\ \text{A} \\ I_{a1} &= \frac{V_{a1}}{Z_1} = \frac{112.78\angle{-0.03^\circ}}{25.30\angle71.57^\circ} = 4.458\angle{-71.60^\circ}\ \text{A} \\ I_{a2} &= \frac{V_{a2}}{Z_2} = \frac{61.62\angle45.88^\circ}{25.30\angle71.57^\circ} = 2.436\angle{-25.68^\circ}\ \text{A} \end{aligned} $$ For reference, the line currents are $I_a = 7.72\angle{-48.16^\circ}$ A, $I_b = 4.15\angle136.40^\circ$ A, $I_c = 2.91\angle30.98^\circ$ A, and the neutral current $I_n = 3I_{a0} = 4.84\angle{-16.69^\circ}$ A. **Answer: $V_0 = 47.77\angle57.63^\circ$, $V_1 = 112.78\angle{-0.03^\circ}$, $V_2 = 61.62\angle45.88^\circ$ V; $I_0 = 1.614\angle{-16.69^\circ}$, $I_1 = 4.458\angle{-71.60^\circ}$, $I_2 = 2.436\angle{-25.68^\circ}$ A.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a star load with series impedance Zs and neutral impedance Zn, what is the zero sequence impedance Z0? A. Zs B. Zs + Zn C. Zs + 3Zn D. 3Zs + Zn Answer: C. Zero sequence currents of all three phases add in the neutral, so Zn appears as 3Zn in series with Zs: Z0 = Zs + 3Zn. (Key point: Zero sequence impedance) 2. Why are Z1 and Z2 equal to Zs only (without Zn)? A. Positive and negative sequence currents sum to zero in the neutral, so no voltage drop appears across Zn B. Zn is shorted for unbalanced voltages C. Zn is only effective for the positive sequence D. Zn is negligible compared with Zs Answer: A. Balanced sets of positive or negative sequence currents carry no neutral current, so Zn has no effect and Z1 = Z2 = Zs. (Key point: Positive/negative sequence impedance) 3. How are the sequence currents obtained from the sequence voltages for this balanced star load? A. I = V times Z for each sequence B. Each sequence voltage is divided by its own sequence impedance, since the sequence impedance matrix is diagonal C. All three voltages are divided by Z0 D. Only the positive sequence is divided by Z1; others are zero Answer: B. A balanced load has a diagonal sequence impedance matrix, so Ia0 = Va0/Z0, Ia1 = Va1/Z1, Ia2 = Va2/Z2. (Key point: Sequence current method) 4. Which expression gives the positive sequence voltage Va1 (a = 1 at 120°)? A. (Va + a^2 Vb + a Vc)/3 B. (Va + Vb + Vc)/3 C. (Va + a Vb + a^2 Vc)/3 D. (Va + a Vb + a Vc)/3 Answer: C. Positive sequence is Va1 = (Va + aVb + a^2Vc)/3; the a^2 and a swap gives negative sequence. 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